More on sequences in groups
Cheng-De Wang Philip A. Leonard
Department of Mathematics Arizona State University Tempe, AZ 85287-1804, USA
Abstract
We bring to conclusion the investigation of three problems about se- quencings for finite groups: the existence of harmonious sequences in dicyclic groups, the R-sequenceability of dicyc1ic groups, and the R- sequenceability of the nonabelian groups of order pq, where p and q are primes.
Introduction
Various types of sequences of the elements of a finite group have been studied in connection with questions in combinatorics. In this article we discuss harmonious sequences and R-sequences, both of which are connected to complete mappings of a finite group G.
Definition 1 A complete mapping of G is a permutation 9 -+ ()(g) of the elements of G such that ¢ : 9 -+ g()(g) is again a permutation of the elements of G. In this case, the mapping ¢ is called an orthomorphism of G.
The idea of a complete mapping was introduced by H. B. Mann [5] and studied later by L. J. Paige [6]. Results related to this notion and to those that follow are discussed in A. D. Keeedwell's recent survey [4].
Definition 2 A group G of order m is called harmonious if its m elements can be listed in a sequence
91, 92,"" 9m such that the products
91g2, 9293, ... ,9m-19m, 9mgl of consecutive elements are all distinct.
These sequences were introduced in [1]; if G has a harmonious sequence, then
e
= (gl, g2,' .. , gm) is a complete mapping of G, expressed as a single m-cycle.Harmonious groups include all groups of odd order, the nontrivial finite abelian groups having noncyclic Sylow 2-subgroups (except for the elementary abelian 2- groups), and the dihedral groups Dn of order 2n whenever n is divisible by 4 or n
=
6m, m odd [1]. In Section 1 we discuss harmoniousness in dicyclic groups.Definition 3 A group of order m is called R-sequenceable if its m elements can be listed in a sequence
_ gl
=
1, 92, ... , gmsuch that the partial products gl, glg2, 9192g3, ... ,glg293 ... gm-l, are all distinct and 91g2g3'" gm-lgm
=
1.If G has an R-sequencing, then there is a complete mapping (J of G such that the corresponding orthomorphism ¢ fixes one element and permutes the remaining elements in a single cycle. The dihedral group Dn is R-sequenceable if and only if n is even [3]. We treat this concept for dicyclic groups in Section 2, and for the nonabelian groups of order pq (p and q prime) in Section 3.
In each of the next three sections, the complete result is designated as a Theorem.
Previous results are indicated as Propositions; the new contributions are clearly indicated in our discussion.
1 Harmoniousness of dicyclic groups
For an nxn matrix M, the (i,j)-entry of M is denoted by M(i,j). For a permutation T of degree n, the collection of n elements {M(i, T(i)), i
=
1,2, ... , n} is denoted by T(M). The dicyclic group Q2n of order 4n is defined byQ2n = (0;,/3: 0;2n = 1,/32
=
o;n,o;/3 = /30;-1).The following proposition was proved in [7].
Proposition 1 Let A and B be two n x n matrices defined by A(i,j) = i
+
j - 2 mod 2n, i,j = 1,2, ... , n B(i,j)=n-i-j+l mod2n, i,j=1,2, ... ,n.Then the dicyclic group Q2n is harmonious if there exist two permutations 7r and (J of degree n such that (J07r is an n-cycle and 7r(A) U(J(B) is a complete set of residues modulo 2n.
Sketch of the proof. Let
f =
(J 0-7r. Let c be an integer with 1 ~ c:::; n. For any fixed integer d, we defineb2i- 1
= -
fi-l(c)+
d, b2i = t-1(c)+
d - 1, a2i-1 = 7r f i- 1 (c) - 1,a2i = 7r fi-l(c)
+
n - 1.We notice that
b2i
+
a2i-l = A(Ji-1(C), 'll-ji-1(C))+
d andb2i+1 - a2i
=
B(7r fi-l(c), fi(c))+
d.Direct calculation shows that
{b2i - b2i -1
+
n : i=
1, 2, ... , n} U {a2i+
a2i-1 : i=
1, 2, ... , n}is a complete set of residues modulo 2n and
{b2i+l - a2i : i
=
1,2, ... , n} U {b2i+
a2i-1 : i=
1,2, ... ,n}=
7r(A) U B(B)+
dwhich is also a complete set of residues modulo 2n by our hypothesis. Therefore, the following sequence is a harmonious sequence of Q2n:
I By using Proposition 1 the following result was proved in [7].
Proposition 2 If n is a multiple of 4 or 6, the dicyclic group Q2n is harmonious.
It remains to deal with the case of n
=
4k+
2. We define two permutations 7rand 8 of degree n by
7r(X) = (
~ ~~!: ~
3k
+
2 x - 2k - 1 14k
+
2 Y+
2k - 1B(y)= k+1
Y
+
2k - 2 y - 2k - 2 y - 2k - 1if 1 :::; x :::; k, if k
+
1 :::; x :::; 2k, if 2k+
1 :::; x :::; 3k+
1, if x=
3k+
2,if 3k
+
3 :::; x :::; 4k+
2, if y = 1,if y
=
2,if 3 :::; y :::; k
+
2, if y=
k+
3,if k
+
4 :::; Y :::; 2k+
3,if 2k
+
4:::; y :::; 3k+
2,if 3k
+
3 :::; y :::; 4k+
2.By the definitions of matrices A and B in Proposition 1, we obtain
7r(A) = {2k, 2k
+
1, 2k+
2, 2k+
3, ... , 4k, 4k+
2, 4k+
3, 4k+
4, 4k+
5, ... , 6k, 6k+
1, 6k+
2}and
8(B) = {O, 1,2,3, ... , 2k - 3, 2k - 2, 2k - 1, 4k
+
1, 6k+
3, 6k+
4, ... , 8k+
1, 8k+
2,8k+
3}.Therefore, 7r(A) U (}(B) is a complete set of residues modulo 2n.
Considering k modulo 3, we find that in two cases () 0 7r is a cycle of length n. If k = 3t, by direct calculation, we have () 0 7r = (14k 4k - 3 4k - 6 4k - 9 ... 3k
+
6 3k+
3 3k+
1 3k 3k - 1 3k - 2 ... 2k+
3 2k+
2 4k+
2 4k - 1 4k - 4 4k - 7 ...3k
+
5 3k+
2 k k - 1 k - 2 k - 3 ... 3 2 4k+
1 4k - 2 4k - 5 ... 3k+
7 3k+
4 k+
1 k+
2 k+
3 ... 2k 2k+
1) which is an n-cycle.If k = 3t
+
2, we have () 0 7r = (14k 4k - 3 4k - 6 4k - 9 '" 3k+
5 3k+
2 k k - 1 k - 2 k - 3 ... 3 2 4k+
1 4k - 2 4k - 5 ... 3k+
6 3k+
3 3k+
1 3k 3k - 1 3k - 2 ...2k+3 2k+2 4k+2 4k-1 4k-4 4k-7 ... 3k+7 3k+4 k+1 k+2 k+3 '" 2k 2k+1) which is a cycle of length n.
Therefore, by Proposition 1 we can state
Proposition 3 lfn
=
12t+2 orn=
12t+10, the dicyclic group Q2n is harmonious.In the remaining we have case n
=
12t+
6, and this is covered by Proposition 2.It is shown in [1] that Q2n is not harmonious if n is an odd integer or n
=
2. Hence, by Propositions 2 and 3 the following is true:Theorem 1 Q2n is harmonious if and only if n is an even integer greater than 2.
It is obvious that a harmonious group may have many harmonious sequences.
We can, for example, give an alternative construction for the case of n
=
8k+
2 as follows.Let 7r and () be permutations of degree n defined by 4k
+
2 +x4k+4+x 2k+ 1 7r(x)
=
4k+
x 4k+2 x - 4k-1 6k+2O(y) = {
~k +
yy - 4k - 1
if 1 ~ x ~ 2k - 1,
if 2k ~ x ~ 4k - 2 and x is even, if x
=
2k+
1,if 2k
+
3 ~ x ~ 4k+
1 and x is odd, if x = 4k,if 4k
+
2 ~ x ~ 8k+
2 and x =I-6k+
2, if x = 6k+
2,if y
=
1,if 2 ~ y ~ 4k
+
2, if 4k+
3 ~ y ~ 8k+
2.By the definitions of matrices A and B in Proposition 1, we have
7r(A) = {4k, 4k
+
1, 4k+
2, ... , 8k - 1, 8k, 8k+
2, 8k+
3, ... , 12k+
2}mod 2n and
(}(B)
=
{O, 1,2, ... ,4k - 1, 8k+
1, 12k+
3, 12k+
4, ... ,16k+
3}mod 2n.Therefore 7r(A) U (}(B) is a complete set of residues modulo 2n.
By direct calculation, we have () 0 7r = (1 2 3 ... 2k - 1 2k 2k
+
3 2k+
2 2k+
5 2k+
4 2k+
7 2k+
6 ... 4k+
1 4k 8k+
2 8k+
1 8k 8k - 1 ... 6k+
3 6k+
2 2k+
1 6k+
1 6k 6k - 1 ... 4k+
2) which is an n-cycle.2 The R-sequenceability of dicyclic groups
The following two propositions were proved in [8].
Proposition 4 Q2n is R-sequenceable if there are integers a2, a3, ... ,a2n-l and b1 ,
b2, ... ,b2n satisfying
(1) 0, a2, a3, ... ,a2n-l are distinct mod 2n, (2) hI, b2; ... ,b2n are distinct mod 2n,
(3) 0, a2, a3 - a2,' .. , an - an-I, bn+l - bn, bn+2 - bn+l , ... , b2n - b2n- l are distinct mod 2n,
(4) bl +an, bl +an+l +n, b2 +an+l? b2 +an+2 +n, b3 +an+2,' .. , bn- l +a2n-l +n, bn + a2n-l, b2n
+
n are distinct mod 2n.Proposition 5 Let A and B be two n x n matrices defined by A(' .)
= {
3n/2+
i+
j - 1 mod 2n if i ~ n/2'l,j 3n/2
+
i+
j mod 2n ifi > n/2 B(' .)= {
n/2+
i+
j - 1 mod 2n if i ~ n/2'l,j n/2
+
i+
j mod 2n ifi > n/2.Then the dicyclic group Q2n is R-sequenceable if there exist two permutations 7r and B of degree n such that 7r 0 (}-1 is a cycle of length n with (}(1)
=
n, and 1l'(A) U B(B) is a complete set of residues modulo 2n.Theorem 2 Q2n is R-sequenceable if and only if n is an even integer greater than 2.
Proof. It was shown in [8] that for n = 2, and for n odd, Q2n is not R-sequenceable, and for n == 0 mod 4, Q2n is R-sequenceable. Thus we assume that n
=
4k - 2 where k > 1 is an integer. We modify the proof of the case of n=
4k in [8] by defining the permutations of 1l' and B by1l'(x) = { X
+
2k - 1 if 1 ~ x ::; 2k - 1, x - 2k+
1 if 2k ~ x ~ 4k - 2,B( )
= {
y+
2k - 2 if 2 ~ y ~ 2k - 1, y =1= k+
1,Y Y - 2k
+
2 if 2k ::;. Y ~ 4k - 3, Y =1= 3k - 2, together with B(l)=
4k - 2, (}(4k - 2) = 1, B(k+
1) = k, and B(3k - 2) = 3k - 1.By the definitions of A and B in Proposition 5, we have 1l'(A)
=
{I, 2, 3, ... ,4k - 2}and (}(B) = {4k - 1, 4k, 4k
+
1, ... , 8k - 4}. Hence 1l'(A) U B(B) is a complete set of residues modulo 2n.Notice k > 1. By direct calculation, we have 1l' 0 B-1
=
(2k - 1 2k - 2 ...k + 1 k 3k 3k + 1 3k + 2 ... 4k - 2 2k 2k + 1 2k + 2 ... 3k - 1 k - 1 k - 2 ... 2 1) which is a cycle of length n. Hence by Proposition 5, Q2n is R-sequenceable. I
By using Proposition 4 we can give a distinct R-sequencing of Q2n from the construction indicated in the proof of Theorem 2. This construction can serve as an alternative proof to Theorem 2 for the case n = 8k
+
2.We define the sequences (1) and (2) in Proposition 4 as follows:
Sequence (1), of 2n - 1 elements, can be given in seven segments, with numbers of elements and rule of construction given by
(i) 8k
+
2 elements: 0, 8k+
1, 1, 8k, 2, 8k - 1, ... , 4k, 4k+
1;(ii) 2k elements: 16k
+
3, 16k+
2, 16k+
5, ... , 14k+
5, 14k+
4;(iii) 2k - 2 elements: 14k
+
1, 14k+
2, 14k - 1, 14k, 14k - 3, 14k - 2, ... ,12k+
5, 12k+
6;(iv) 2k elements: 12k
+
3, 12k+
2, 12k+
1, 12k, ... , 10k+
4;(v) The single element 14k
+
3;(vi) 2k
+
1 elements: 10k+
3, 10k+
2, ... , 8k+
3;(vii) The final element 12k
+
4.Similarly sequence (2), of 2n elements, is given by (i) 2k
+
1 elements: 8k+
1, 8k, 8k - 1, ... , 6k+
1;(ii) 2k - 2 elements: 6k - 2, 6k - 1, 6k - 4, 6k - 3, 6k - 6, 6k - 5, ... , 4k
+
2, 4k +3;(iii) 2k elements: 4k, 4k - 1, 4k - 2, ... , 2k
+
1;(iv) the single element 6k;
(v) 2k
+
1 elements: 2k, 2k - 1, 2k - 2, ... , 1, 0;(vi) the single element 4k
+
1;(vii) 8k
+
2 elements: 12k+
3, 12k+
2, 12k+
4, 12k+
1, 12k+
5, 12k, 12k+
6, 12k - 1, ... , 16k+
3, 8k+ 2.In the following examples, semicolons separate the segments in the listing of sequence elements. k = 1, one segment of each sequence is vacuous (noted by a repeated semicolon).
When k
=
1, so that n=
8k+
2=
10, the sequences are(11): 0, 9, 1, 8, 2, 7, 3, 6, 4, 5; 19, 18;; 15, 14; 17; 13, 12, 11; 16, and (21): 9, 8, 7;; 4, 3; 6; 2, 1, 0; 5; 15, ~4, 16, 13, 17, 12, 18, 11, 19, 10.
When k = 2, so that n = 8k
+
2 = 18, the sequences are(12): 0, 17, 1, 16, 2, 15, 3, 14, 4, 13, 5, 12, 6, 11, 7, 10; 8, 9; 35, 34, 33, 32; 29, 30;
27,26,25,24; 31; 23, 22,21,20,19; 28,and
(22): 17, 16, 15, 14, 13; 10, 9; 8, 7, 6, 5; 12; 4, 3, 2, 1, 0; 9; 27, 26, 28, 25, 29, 24, 30, 23, 31, 22, 32, 21, 33, 20, 34, 19, 35, 18.
3 R-sequenceability of groups of order pq
First we state an alternative definition of R-sequenceability. It is easily seen that this definition is equivalent to Definition 3.
Definition 4 (a) A group G of order m is called R-sequenceable if its noniden- tity elements can be listed in a sequence
such that
are all distinct.
(b) If, further, gl = g2gm-l = gm-lg2, G is called R*-sequenceable.
The nonabelian group G of order pq where p, q are primes with q
==
1 (mod p) is defined, using r such that rP==
1 (mod q), r1=
1 (mod q), as follows:G
=
{(u, v)I
u E Z/pZ, v E Z/qZ}with multiplication defined by
(u, v)(x, y) = (u
+
x, vrX+
y).The following result was proved by Keedwell [3].
Proposition 6 The nonabelian group of order pq is R-sequenceable if p has 2 as a primitive root.
This result can be extended, essentially by means of a modification of the direct product theorem in [2].
Theorem 3 The nonabelian group of order pq is R-sequenceable.
Proof. In view of Proposition 6 we assume p > 5. The (additive) cyclic group Z/pZ is R*-sequenceable [2]; let gl, ... , gp-l be an R*-sequencing with gl = g2
+
gp-l.In order to list the nonidentity elements of G, we introduce p x q matrices A and B, defined over Z / pZ and Z / qZ, respectively. Entries all and bl1 are left blank; group elements are given as (aij, bij ), and are sequenced by reading column-by-column from A and B.
Let aij = gl for i = 1, 2 ::; j ::; (q
+
1)/2, and for i = 2, 1 ::; j ::; (q+
1)/2. For i=
1,2 and j > (q+
1)/2, let aij = 0. For i > 2 and 1 ::; j ::; q, let aij=
gi-l' The definition of B uses the nonzero element c=
_rgl-gp-2-g2 of Z / qZ; all arithmetic is in Z/qZ. For j 2:: 2, let b1j = b3j = (j - 1) . c. Complete row 3 by letting b31 = 0, and define row 2 by letting b2j = -b3j , 1 ::; j ::; q. For 4 ::; i ::; P - 1 (except as noted below), row i is the q-tuple (0, q - 1, q - 2, ... ,2,1). The final rowis (rk, 2rk, .. . , (q -l)rk, 0), where k = 9p-l - 9p-2 mod p. (A modification is needed when crg3 - g2 == -1, or equivalently, when 93 - 92
+
9p-l - 9p-2 == 0 mod p. In this case, row 4 of B is defined by the q-tuple (0,1,2, ... , q - 1), affecting rows 4 and 5 of B' below.)To verify that (aij, bij ) form the desired R-sequencing of G, one must study the matrices A' and B' with a~1 and b~1 blank and other entries defined by
(
I .. b~.)
= {
(ai-1,j, bi_1,j)-l(aij, bij ) if i > 1,azJ ' ZJ ( ap,j-b p,j-1 b ) -1 ( aij, ij b ) ' 1 f . - 1 't - an d' J > , 1
together with (a~lb~l) = (apqbpq)-1(a21b2d. Thus the (a~j' b~j) list the products re- quired by Definition 3.
The verification that the listed products are distinct is aided by the following observations:
(i) The matrix A' has a particularly simple form, namely,
91 - 9p-l 91 - 9p-l 92 - gl 92 - gl
91 - 9p-1 0 0 0 0
92 - 91 92 - 91 92 - 91 91 - 9p-1 91 - 9p-l
93 - g2 93 - 92 93 - 92
gp-1 - gp-2 9p-l - 9p-2 9p-1 - gp-2
so that, from row 4 onwards, it suffices to verify that the entries in each row of B' are distinct.
(ii) q
=
2hp+ 1 and rP=
1, so that none of rg2 or r92 - g1 , r g4 - g3 , ... , r9p-2-9p-3 can be -1.(iii) The analysis of row 1 and row 3 is made easy by observing the two equalities c - rgl-gp-2
=
crg2+
c, c - r-gp-2=
crg2 - g1+
c.(iv) The last row of B' is given by (rk, 3rk, ... , (q - 2)rk, 0, 2rk, 4rk ... , (q -l)rk).
(v) If crg3-g2 = -1, then row 4 of B' as "generally" defined consists entirely of zeros. The modification noted above corrects this difficulty so that the affected rows
of B' have distinct entries. I
Remark: It should be pointed out that the R-sequencing of the previous result is, in fact, an R*-sequencing because (91,0)
=
(g2, 0)(9p-l, 0) = (9p-b 0)(g2, 0).It is useful to have an example for the construction of the R-sequencing in The- orem 3. We include the smallest example for which the prime p does not have 2 as a primitive root.
Example. The nonabelian group G of order 203 has p
=
7 and q=
29. We use the R*-sequencing 3, 2, 5, 6, 4, 1 of Z/7Z, so that A is as follows:-3333333333333300000000000000 33333333333333300000000000000 22222222222222222222222222222 55555555555555555555555555555 66666666666666666666666666666 44444444444444444444444444444 1 III 1 1 1 1 1 1 1 1 III III 1 1 1 1 III 1 III
Use r
=
7 (which implies c = 6). Since g3 - g2+
gp-l - gp-2==
0 mod 7, we modify row 4 and obtain the following matrix as B:- 06 12 18 24 01 07 13 19 25 02 08 14 20 26 03 09 15 21 27 04 10 16 22 28 05 11 17 23
o
23 17 11 05 28 22 16 1004 27 21 15 09 03 26 20 1408 02 25 19 13 0701 24 18 12 06o
06 12 18 24 01 07 13 19 25 02 08 14 20 26 03 09 15 21 27 04 10 16 22 28 05 11 17 23o
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28o
28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 09 08 07 06 05 04 03 02 01o 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 09 08 07 06 05 04 03 02 01 23 17 11 05 28 22 16 10 04 27 21 15 09 03 26 20 14 08 02 25 19 13 07 01 24 18 12 06 0
By the definitions of the elements (a~j' b~j) we have A' as follows:
-2222222222222266666666666666 20000000000000000000000000000 66666666666666622222222222222 33333333333333333333333333333 1 111 1 111 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 555 5 555 5 555 5 5 5 5 5 5 5 5 5 5 5 555 5 5 5 5 44444444444444444444444444444 and, finally, B' is as follows:
- 10 20 01 11 21 02 12 22 03 13 23 04 14 24 20 02 13 24 06 17 28 10 21 03 14 25 07 18
o
17 05 22 10 27 15 03 20 08 25 13 01 18 06 23 11 28 16 04 21 09 26 14 02 19 07 24 12o
11 22 04 15 26 08 19 01 12 23 05 16 27 09 05 15 25 06 16 26 07 17 27 08 18 28 09 19o 02 04 06 08 10 12 14 16 18 20 22 24 26 28 01 03 05 07 09 11 13 15 17 19 21 23 25 27
o
21 13 05 26 18 10 02 23 15 07 28 20 12 04 25 17 0901 22 14 06 27 19 11 03 24 16 08o 15 01 16 02 17 03 18 04 19 05 20 06 21 07 22 08 23 09 24 10 25 11 26 12 27 13 28 16 23 11 28 16 04 21 09 26 14 02 19 07 24 12 0 17 05 22 10 27 15 03 20 08 25 13 01 18 06 The listing of the elements (aij, bij ) by columns gives the indicated R* -sequencing of G, as can be checked by examining the (a~j' b~j)'
References
[1] R. Beals, J.A. Gallian, P. Headley and D. Jungreis, Harmonious groups, J.
Combinatorial Theory A 56 (1991), 223-238.
[2] R.J. Friedlander, B. Gordon, and M.D. Miller, On a group sequencing problem of Ringel, Congr. Numer. 21 (1978), 307-321.
[3] A.D. Keedwell, On R-sequenceability and Rh-sequenceability of groups, Ann.
Discrete Math. 18 (1983), 535-548.
[4] A.D. Keedwell, Complete mappings and sequencings of finite groups, in "eRC Handbook of Combinatorial Designs" (C.J. Colbourn and J.H. Dinitz eds.), CRC Press,(1996), 246-253.
[5] H.B. Mann, The construction of orthogonal Latin squares, Ann. Math. Statistics 13 (1942), 418-423.
[6] L.J. Paige, Complete mappings of finite groups, Pacific J. Math. 1 (1951),111- 116.
[7] C-D. Wang, On the harmoniousness of dicyclic groups, Discrete Math. 120 (1993), 221-225.
[8] C-D. Wang, On the R-sequenceability of dicyclic groups, Discrete Math. 125 (1994), 393-398.
(Received 21/5/99)
